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FIN HEAT TRANSFER CLASS NOTES FOR MECHANICAL ENGINEERING

FIN HEAT TRANSFER CLASS NOTES FOR

MECHANICAL ENGINEERING  

Fin is a solid extended surface  or a combined conduction convection system. Heat is transferred  by Conduction within the solid and by convection from the fin surface in a PERPENDICULAR  direction to that of conduction. Finally heat is lost to the surroundings. Fins increase the rate of heat transfer by increasing the surface area without increase of primary surface area. This increased surface area reduces the convective resistance and increases the rate of heat transfer economically. The fins should have large  surface area per unit volume (>700 m2/m3) with small gas flow passages with natural laminar flow for increased rate of heat transfer. It increases the heat exchange between two fluids which are separated by a sold surface as  in home radiators and cooling system in cars. 

PURPOSE OF A FIN:  To increase heat transfer by increasing the surface area without increase of primary surface area. FIN involves both (CONDUCTIVE + CONVECTIVE) HEAT TRANSFER. . Finally it loses heat to the atmosphere.  

HOW FINS INCREASE HEAT TRANSFER:  Convection equation  q =h A dT   Watts.

Undoubtedly heat transfer can be increased by increasing the

(i)  temperature difference

(ii)   the convection heat transfer coefficient

(iii)  the contact surface area of the object

              ΔT can’t be increased because of fin base temperature and environment temperature are normally fixed. Convective heat transfer coefficient can be increasing by using a fan to blow air over the surface. However , this has not been practically possible to increase ‘h’.  Thus it is not feasible to change the first two options.  Therefore, increasing the surface area seems to be the only choice possible. ‘This can be easily and economically achieved by  fins.

TYPE OF FINS / SHAPES OF FINS

(i)Fins of constant area of cross section like rectangular, circular, square and elliptical fins

(ii) Fins with varying area of cross section like triangular, trapezoidal , parabolic and conical

(iii) Plate fins and wound fins from the two types above

Fig. Types of fins

 PURPOSE OF DESIGNING A FIN

Fin should be most effective. It should be able to transfer many times the heat as compared without fins. It should also be cost effective.

Write the assumption used in the analysis of heat transfer through fins.

 FIN PARAMETER

 Fin parameter is represented by the letter ‘m’. It is given by the relation given below

m = (h P/k Ac)0.5

Where

h surrounding medium convective HT coefficient

P is the perimeter of the fin

k  thermal conductivity of fin material

Ac cross sectional area of the fin if rectangular, Ac =b x t

Q. FIN EFFICIENCY FOR A SINGLE FIN

It is applicable for a single fin.

 Ratio of the actual rate of heat transfer to the maximum rate of heat transfer through the fin. Maximum heat transfer occurs if the entire fin were at the base temperature.

The fin efficiency is ηf = actual q.fin/ q.max with base temperature all along the length

ηf = (PhAck)1/2(Tb–T )/hPL (Tb–T )

     = (kAc/hP)1/2(1/L)

     =1/mL

   Finally ηf =1/mL

Fin efficiency will always be less than one. This is

because assuming the temperature throughout

the fin is at the base temperature would increase

the heat transfer rate.

PRACTICAL APPLICATIONS OF FINS

(i)  Electric motors

(ii) Car radiators

(iii) Pumps

(iv) Compressors

(v)  Internal Combustion engines

(vi) Transformers

(vii) Condensers

(viii) Evaporators

(ix)  Air craft engines

(x)  Economizers of steam power plants

(xii) In nature, the ears of Jackrabbits and Fennec Foxes act as fins to release heat from within their bodies.

Fin Performance Parameters

Description of Fin performance in three different ways

(i)   Effectiveness of a single fin

(ii)  Efficiency of a single fin

(iii)Overall efficiency of an array of fins

 FIN EFFECTIVENESS OF A SINGLE FIN

It is applicable for a single fin. It is the ratio of heat transfer with fin to the rate to the heat transfer without fin. Thus it is greater than 1. Practically it should be at least greater than 2.3.

The formula for a fin of infinite length

Єf = q. with fin/ q.without fin= (P h Ac k)1/2(Tb–T)/ h Ac (Tb–T)

= (Pk/h Ac)1/2

Єf = (Pk/h Ac)1/2

Where Ais the fin cross-sectional area

For a rectangular fin, P = 2B + 2 t and Ac = B t

Where B is width of fin and t is the thickness of fin

Now if the fin is thin, Neglect 2t as compared to 2B

Therefore P/Ac = 2B/B t=2/t

Єf = (2k/ht)1/2 = (1/Bi)0.5

Biot number, Bi for a fin

Less value of Biot number is desirable for more effectiveness.

Biot number will be less if h is small i.e. gas environment all around the fin.

Further, Biot number will be less if thermal conductivity k of fin material is high.

Biot number will be less if the thickness of fin is less i.e. fin is thin.

ASSUMPTIONS USED IN FIN ANALYSIS

  1. One-dimensional steady state heat conduction

  2. Homogeneous and isotropic fin material, k is constant

  3. No internal heat generation

  4. Uniform cross-sectional area

  5. Uniform convective heat transfer coefficient h

  6. Negligible contact thermal resistance

  7. Negligible radiation

  8. Always calculate Rate of heat transfer at the base of the fin

NOMENCLATURE OF FIN

  1. Base of the fin or root of the fin: Where fin connects to the main body

  2. Tip of the fin—The other free end of the fin

  3. Ac is cross sectional area of the fin = b x t for a rectangular fin

  4. As is the surface area of the fin= perimeter x length= (2b+2t) L

Fig. Details of a rectangular fin

DERIVATION OF BASIC DIFFERENTIAL EQUATION FOR FINS OF CONSTANT AREA OF CROSS SECTION

General partial equation for a fin is

d2θ/dx2 –m2 θ=0

where parameter m = (hP/kAc)0.5

θ is the temperature difference

DERIVATION

Using above assumptions, apply the Law of conservation of energy for a differential element of the fin. Consider two sections at distances x and x+ dx from the fin base. Note conduction takes place through the cross-sectional area.  Convection takes place on the surface area and perpendicular to the conduction direction.

 conduction at distance ‘x’  = conduction at (x+dx) + convection through length dx

qx = qx+dx + dqconv

qx+dx =qx+ (dqx/dx) dx

From Fourier equation

Firstly qx= –kAc (dT/dx)

Secondly qx+dx = –kAc (dT/dx) + d(–kAc (dT/dx) /dx) dx

Simplify qx+dx = –kAc (dT/dx)  — kAcd(dT/dx) /dx) dx

qx+dx = –kAc (dT/dx)  — kAc d2T/dx2dx

dqconv = h dAs (Tx –T)

dAs = Pdx is the surface area of the differential element

Where P is fin perimeter = (2b+2t) for a rectangular fin

Substituting

0 = -k Ac (d2T/dx2 )dx  + h Pdx (Tx –T)

 Since Ac is constant, it becomes 

d2T/dx2 =(hP/kAc) θ        where (Tx –T) =θ  &  d2T/dx2  =d2θ/dx

d2θ/dx2 –m2 θ=0

Where fin parameter m = (hP/kAc)0.5

It is a second order differential equation.

θ is the temperature difference

The solution will be

θ(x) = C1emx + C2e-mx = C3cosh mx+C4 Sinh mx

Both exponential and hyperbolic solutions are equivalent.

Find the constants C1, C2, C3 & C4  from the boundary conditions.

h surrounding medium convective HT coefficient

K thermal conductivity of fin material

Ac cross sectional area of the fin if rectangular=b x t

Finally the equation is  

d2θ/dx2 –m2 θ=0  

It is a second order differential equation.

Where fin parameter m = (hP/kAc)0.5

ϴ is the temperature difference θx = Tx— Ta

The solution of the differential equation will be

ϴ(x) = C1emx + C2e-mx = C3cosh mx+Csinh mx

Both exponential and hyperbolic solutions are equivalent solutions.

This is a common equation for the three cases.

Find the constants C1& C2  (or C3 and C4)from the boundary conditions.

h surrounding medium convective HT coefficient

k  thermal conductivity of fin material

Ac cross sectional area of the fin if rectangular, Ac =b x t

THREE DIFFERENT CASES OF FIN ANALYSIS

Fig. Different cases of fin

Case 1: Infinitely long fin:  To find Rate of heat transfer and temperature distribution for a very long fin

Boundary conditions   (a) at x=0, ϴ=ϴ= Tb-Ta

(b) At x = ∞, T =Taand thus ϴ= Ta-Ta =0

(1) Put  x=0,  C1 +C2  = ϴ0

(ii) Put x =∞, we get C1 e+ C2e-∞ =0 = C1 e

therefore C1 =0 and then C2  = ϴ0

 ϴ (x) =ϴ0 e-mx

ϴ(x)/ ϴ0    = e-mx

(Tx –Ta)/( Tb –Ta) = e–mx

At distance x from the fin base, Find the temperature. Thus the temperature distribution becomes   known.

Rate of heat transfer at the base of the fin will be

q. =(hPkAc)0.5( Tb –Ta)

Case2: Insulated fin tip: To find Rate of heat transfer and temperature distribution for insulated fin tip

The boundary conditions will be

(1)   at x=0, ϴ=ϴ= Tb-Ta

(2)   at x =L, dϴ/dx=0

Temperature distribution

(Tx –Ta)/( Tb –Ta) =Cosh m(L-x)/Cosh mL

Rate of heat transfer at the base of the fin will be

q. =(hPkAc)0.5( Tb –Ta) Tanh mL

 Case 3: Real finite fin: To find Rate of heat transfer and temperature distribution for a fin of finite length:

Boundary conditions are

(1)   at x=0, ϴ=ϴ= Tb-Ta

(2)   at x=L, the rate of heat transfer by convection on area Ais equal to the rate of conduction through the area Ac

Temperature distribution

(Tx –Ta)/( Tb –Ta) =

[Cosh m(L-x) + (h/km)Sinh m(L—x)]/[Cosh mL + (h/km) Sinh mL]

Rate of heat transfer at the base

q. =(h P k Ac)0.5( Tb –Ta) Tanh mL [Tanh mL + h/km]/ [ 1 + (h/km) Tanh mL]

From the above three cases , case I and case II are much simpler than case III. Fins in actual use will correspond to case III.

When to use analysis of case I and case II for case III?

(i) Use equations of case I for case III when mL=5 with maximum error in calculations has been found to be < 1 % which is negligible.

(ii) Use equations of case II for case III, using corrected length in place of L in case II

Corrected length = actual length + 1/2 thickness of fin

With corrected length, maximum error in calculations has been found to be < 1 % which is negligible.

Fig. Heat loss in 3 cases of fins

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